If the bond energies of $\mathrm{H}-\mathrm{H},\mathrm{Br}-\mathrm{Br}$ and $\mathrm{H}-\mathrm{Br}$ are 433,192 and $364\mathrm{ }\mathrm{k}\mathrm{J}{\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$ respectively, then $\Delta {\mathrm{H}}^0$ for the reaction ${\mathrm{H}}_{2\left(g\right)}+{\mathrm{Br}}_{2\left(g\right)}⟶2\mathrm{HBr}$

 The bond between hydrogen and bromine molecules is broken and bond formation takes place between hydrogen and bromine atoms to form $\mathrm{HBr}$

One mole of hydrogen and bromine reacts to give 2 moles of $\mathrm{HBr}$

 $\Delta H^0=\Delta H_{H-H}+\Delta H_{Br-Br}-\Delta H_{H-Br}$

 $\Delta H^0=433+192-2\left(364\right)$

 $\Delta H^0=625-728=-103$

Hence correct answer will be (D).